The [modified Bessel function of the 1st kind][1] $I_0$ is defined by
$$
I_0(z)=\frac1\pi\int_0^{2\pi}e^{z\cos\theta}\,d\theta
$$
and arises, among other places, in the probability density function of a noncentral $\chi^2(2)$ random variable.

The [error function][2] is defined by
$$
\text{erf}(z)=\frac2{\sqrt\pi}\int_0^z e^{-t^2}dt
$$
and is closely related to the cumulative distribution function for the standard normal distribution.

I believe these are not [elementary functions][3], but
>is either $I_0$ or $\text{erf}$ elementary "**relative to**" the other?

That is, if we were to add one of them to our repertoire of "elementary" functions, would the other one become "elementary"?


----------
**Edit:** @Suvrit mentioned that with the confluent hypergeometric function $_1F_1=M$,
$$
\text{erf}(z)=\frac2{\sqrt\pi} M\left(\frac12,\frac32,-z^2\right)\cdot z\qquad\text{and} $$
$$
I_0(z)= M\left(\frac12,1,2z\right)e^{-z},\qquad\text{where}
$$
$$M(1/2;b;z)=\sum_{n=0}^\infty \frac{(1/2)^{(n)}z^n}{b^{(n)}n!}\qquad\text{and}$$
$$
a^{(n)}=a(a+1)\cdots (a+n-1)
$$
so that $a^{(2)}=(a+1)a$ etc. In particular
$$(1/2)^{(n)} = \frac12 \frac32 \frac52 \dots = \frac{(2n)!}{n!4^n}\qquad\text{and}$$
$$\frac{(1/2)^{(n)}}{(3/2)^{(n)}}=\frac{\frac12\frac32\frac52\dots\frac{2n-1}{2}}{\frac32\frac52\frac72\dots\frac{2n+1}{2}}=\frac{1/2}{(2n+1)/2}=\frac1{2n+1}.$$
So the question can be rephrased as:
>are the following functions elementary relative to eachother?
\begin{align}
f(z):=M\left(\frac12,\frac32,-z^2\right)&=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}\frac{z^{2n}}{n!},&\qquad\text{and}\\
g(z):=M\left(\frac12,1,2z\right)&=\sum_{n=0}^\infty \frac{(2n)!}{n!2^n}\frac{z^n}{n!}.&
\end{align}

  [1]: http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html
  [2]: http://en.wikipedia.org/wiki/Error_function
  [3]: http://en.wikipedia.org/wiki/Elementary_function